# Show that $\tau$ is strictly finer than the euclidean topology.

Let $$A=\lbrace \frac{1}{n} : n \in \Bbb{N} \rbrace$$, $$H=\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$$. Let $$\tau$$ be the topology on $$\Bbb{R}$$ generated by the basis $$H$$.

First, I want to ask about $$\tau$$. How can I find it? I know $$H$$ is a basis for $$\tau$$, so any element of $$\tau$$ written as a union of element of $$H$$. Does this mean that $$\tau = \bigcup \lbrace \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace \rbrace$$ ?

I have another question about $$H$$. I think $$\lbrace (a,b) -A \rbrace \subset (a,b)$$, so is $$\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace= \lbrace (a,b) \rbrace$$ ?

Edit:

Show that $$\tau$$ is strictly finer than the Euclidean topology.

Let $$\tau '$$ is the Euclidean topology in $$\Bbb{R}$$.We know that $$B=\lbrace (a,b):a,b \in \Bbb{R} \rbrace$$ is a basis of $$\tau '$$, so any open set in $$\tau '$$ written as union of elements of $$B$$, then and from the definition of $$H$$, it is obvious that $$B\subset H$$, hence $$\tau ' \subset \tau$$. This means that $$\tau$$ is finer than the Euclidean topology. We have that $$C=(-1,1)\setminus A$$ is open in $$\tau$$, but is not in the Euclidean topology, because $$0$$ is element of $$C$$ but not an interior point to $$C$$. This means that $$\tau$$ is finer than the Euclidean topology. 0

Find $$A'$$. (The limit points of $$A$$).

Let $$x \in \mathbb{R}$$.

• If $$x<0$$, there exists a neighborhood of $$x$$, which is $$(x-1,0)$$, such that $$(x-1,0)\cap A=\emptyset$$. Thus $$x\notin A’$$ for all $$x<0$$.
• If $$x=0$$, there exists a neighborhood of $$0$$, which is $$V=(-1,1)\setminus A$$, such that $$V\cap A=\emptyset$$. Thus $$0\notin A’$$.
• If $$0, there exists a neighborhood of $$x$$, which is $$U=(0,1)\setminus A$$, such that $$U\cap (A\setminus \lbrace x \rbrace)=\emptyset$$. Thus $$x\notin A’$$.

• If $$x=1$$, we have that $$(\frac{3}{4},2)$$ is a neighborhood of $$1$$, and $$(\frac{3}{4},2)\cap (A\setminus \lbrace 1 \rbrace )=\emptyset$$. Hence $$1 \notin A’$$.

• If $$x>1$$, there exists a neighborhood of $$x$$, which is $$(\frac{x+1}{2},x+1)$$, and $$(\frac{x+1}{2},x+1) \cap A=\emptyset$$. Then $$x\notin A’$$.

As a result, $$A’=\emptyset$$.

No, $$\bigcup \lbrace \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace \rbrace = \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$$

{(a,b) - A} is not a subset of (a,b).
(a,b) - A is a subset of (a,b).
Also $$\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$$ = { (a,b), (a,b)-A }

The topology is all sets of the form
$$\cup$${ (a,b) : (a,b) in S} $$\cup$$ $$\cup$${ (a,b) - A : (a,b) in S }
where S and T are subsets of R×R.

• I edit my answer, can you see it please? Thanks. – Dima Feb 22 '19 at 18:10

You don't have to "find" $$\tau$$, i.e. completely characterize all the $$\tau$$-open sets. All you have to do is exhibit just one set that is open w.r.t. $$\tau$$ but isn't open w.r.t. to the standard Euclidean topology. For this instance it might be easier to exhibit a $$\tau$$-closed set that isn't Euclidean-closed, which is equivalent.

• I edit my answer, can you see it please? Thanks. – Dima Feb 22 '19 at 18:10